Chapter 13 Physics Homework

CHAPTER 13 HOMEWORK 1. Consider the system shown in the figure.(Figure 1) It consists of a block of mass m attached to a spring of negligible mass and force constant k. The block is free to move on a frictionless horizontal surface, while the left end of the spring is held fixed. When the spring is neither compressed nor stretched, the block is in equilibrium. If the spring is stretched, the block is displaced to the right and when it is released, a force acts on it to pull it back toward equilibrium. By the time the block has returned to the equilibrium position, it has picked up some kinetic energy, so it overshoots, stopping somewhere on the other side, where it is again pulled back toward equilibrium. As a result, the block moves back and forth from one side of the equilibrium position to the other, undergoing oscillations. Since we are ignoring friction (a good approximation to many cases), the mechanical energy of the system is conserved and the oscillations repeat themselves over and over. The motion that we have just described is typical of most systems when they are displaced from equilibrium and experience a restoring force that tends to bring them back to their equilibrium position. The resulting oscillations take the name of periodic motion. An important example of periodic motion is simple harmonic motion (SHM) and we will use the mass-spring system described here to introduce some of its properties. PART A: Which of the following statements best describes the characteristic of the restoring force in the spring-mass system described in the introduction? The restoring force is constant. The restoring force is directly proportional to the displacement of the block. The restoring force is proportional to the mass of the block. The restoring force is maximum when the block is in the equilibrium position. PART B: As shown in the figure(Figure 2), a coordinate system with the origin at the equilibrium position is chosen so that the x coordinate represents the displacement from the equilibrium position. (The positive direction is to the right.) What is the initial acceleration of the block, a0, when the block is released at a distance A to the right from its equilibrium position? F= -kA From Newton’s 2nd law (F= ma) 𝑎 = 0 −𝑘𝐴 𝑚 CHAPTER 13 HOMEWORK PART C: What is the acceleration 𝑎 1 of the block when it passes through its equilibrium position? x=0 F= 0 𝑎 1 = 0 HINT: Your results from Parts B and C show that the acceleration of the block is negative when the block has undergone a positive displacement. Then, the acceleration's magnitude decreases to zero as the block goes through its equilibrium position. What do you expect the block's acceleration will be when the block is to the left of its equilibrium position and has undergone a negative displacement? PART D: Select the correct expression that gives the block's acceleration at a displacement x from the equilibrium position. Note that x can be either positive or negative; that is, the block can be either to the right or left of its equilibrium position. 𝑎 =− 𝑘 𝑚 𝑥 HINT: Whether the block undergoes a positive or negative displacement, its acceleration is always opposite in sign with respect to displacement. Moreover, the block's acceleration is not constant; instead, it is directly proportional to displacement. This is a fundamental property of simple harmonic motion. PART E: Explain The magnitude of the block's acceleration reaches its maximum value when the block is at either its rightmost or leftmost position. PART F: Explain The speed of the block is zero when it is at either its rightmost or leftmost position. PART G: Explain The speed of the block reaches its maximum value when the block is in the equilibrium position. PART H: Because of the periodic properties of SHM, the mathematical equations that describe this motion involve sine and cosine functions. For example, if the block is released at a distance A from its equilibrium position, its displacement x varies with time t according to the equation x=Acosωt, where ω is a constant characteristic of the system. If time is measured is seconds, ω must be expressed in radians per second so that the quantity ωt is expressed in radians. Use this equation and the information you now have on the acceleration and speed of the block as it moves back and forth from one side of its equilibrium position to the other to determine the correct set of equations for the block's x components of velocity and CHAPTER 13 HOMEWORK acceleration, vx and ax, respectively. In the expressions below, B and C are nonzero positive constants. vx=−Bsinωt, ax=Ccosωt vx=Bcosωt, ax=Csinωt vx=−Bcosωt, ax=−Ccosωt vx=−Bsinωt, ax=−Ccosωt HINT: Further calculations would show that the constants B and C can be expressed in terms of A and ω . 2. As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a horizontally moving block attached to a spring. Note that, since the gravitational potential energy is not changing in this case, it can be excluded from the calculations. For such a system, the potential energy is stored in the spring and is given by 𝑈 = 1 2 𝑘𝑥 2 , where k is the force constant of the spring and x is the distance from the equilibrium position. The kinetic energy of the system is, as always, 𝐾 = 2 1 2 𝑚𝑣, where m is the mass of the block and v is the speed of the block. We will also assume that there are no resistive forces; that is, E=constant. Consider a harmonic oscillator at four different moments, labeled A, B, C, and D, as shown in the figure (Figure 1). Assume that the force constant k, the mass of the block, m, and the amplitude of vibrations, A, are given. Answer the following questions. PART A: Which moment corresponds to the maximum potential energy of the system? A B C D PART B: Which moment corresponds to the minimum kinetic energy of the system? A B C D CHAPTER 13 HOMEWORK PART C: Consider the block in the process of oscillating. If the kinetic energy of the block is increasing, the block must be at the equilibrium position. at the amplitude displacement. moving to the right. moving to the left. moving away from equilibrium. moving toward equilibrium PART D: Which moment corresponds to the maximum kinetic energy of the system? A B C D PART E: Which moment corresponds to the minimum potential energy of the system? A B C D PART F: At which moment is K=U? A B CHAPTER 13 HOMEWORK C D PART G: Find the kinetic energy K of the block at the moment labeled B. 𝐾 = 3 8 𝑘𝐴 2 3. After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0 cm. The explorer finds that the pendulum completes 97.0 full swing cycles in a time of 136 s. PART A: What is the magnitude of the gravitational acceleration on this planet? L= 53.0 cm = 0.53 m n= 97.0 t= 136 s T= time period of the pendulum g= magnitude of gravitational acceleration on the plant T= t/n T= 136/97 T = 1.4 sec T = 2π√L/g 1.4= 2(π)√0.53/g 𝑔 𝑝𝑙𝑎𝑛𝑒𝑡 = 10.6 m/s² 4. The position of a mass oscillating on a spring is given by x=(7.1cm)cos[2πt/(0.76s)] . You may want to review (Pages 421 - 424). PART A: What is the frequency of this motion? x= (7.1 cm) cos [ 2πt/(0.76s)] x=Acos[2πt/T] Period =T Frequency f = 1/T f= 1/ 0.76s 𝑓 = 1. 3 𝐻𝑧 PART B: When is the mass first at the position x=−7.1cm ? CHAPTER 13 HOMEWORK t= T/2 t= 0.76/2 t = 0.38 s 5. A freshly caught catfish is placed on a spring scale, and it oscillates up and down with a period of 0.2 s. PART A: If the spring constant of the scale is 2040 N/m, what is the mass of the catfish? Time period, t= 2pi *sqrt (m/k) 0.2=2*3.14* sqrt (m/2040) sqrt(m/2040) = 0.2/(2*3.14) m= 2.07 kg 6. You rev your car's engine to 2000 rpm (rev/min). PART A: What is the period of the engine? 𝑤 = 2π(2000𝑟𝑝𝑚( 1𝑟𝑝𝑠 20 𝑟𝑝𝑚 )) = 𝑇 = 3. 0 𝑥10 −2 𝑠 PART B: What is the frequency of the engine? f = 33 Hz PART C: If you change the period of the engine to 0.045 s, how many rpms is it doing? CHAPTER 13 HOMEWORK f = 1300 rpm 7. IP A 3.0 kg mass on a spring oscillates as shown in the displacement-versus-time graph in the figure(Figure 1). PART A: Referring to the graph, at what times between t=0 and t=6.0s does the mass experience a force of maximum magnitude? t= 1.0, 3.0, 5.0 s PART B: EXPLAIN If displacement is large then force is maximum PART C: Calculate the magnitude of the maximum force exerted on the mass. 𝐹𝑚𝑎𝑥 = 3. 7 𝑁 PART D: At what times shown in the graph does the mass experience zero force? CHAPTER 13 HOMEWORK t= 0, 2.0, 4.0, 6.0 s PART E: EXPLAIN force is zero PART F: How much force is exerted on the mass at the time t=0.50s? F= -2.6 N